Hindawi Publishing Corporation Boundary Value Problems Volume 2011, Article ID 516481, 17 pages doi:10.1155/2011/516481 Research Article Existence of Solutions to Nonlinear Langevin Equation Involving Two Fractional Orders with Boundary Value Conditions Anping Chen1, and Yi Chen2 Department of Mathematics, Xiangnan University, Chenzhou, Hunan 423000, China School of Mathematics and Computational Science, Xiangtan University, Xiangtan, Hunan 411005, China Correspondence should be addressed to Anping Chen, chenap@263.net Received 30 September 2010; Revised 21 January 2011; Accepted 26 February 2011 Academic Editor: Kanishka Perera Copyright q 2011 A Chen and Y Chen This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited We study a boundary value problem to Langevin equation involving two fractional orders The Banach fixed point theorem and Krasnoselskii’s fixed point theorem are applied to establish the existence results Introduction Recently, the subject of fractional differential equations has emerged as an important area of investigation Indeed, we can find numerous applications in viscoelasticity, electrochemistry, control, electromagnetic, porous media, and so forth In consequence, the subject of fractional differential equations is gaining much importance and attention For some recent developments on the subject, see 1–8 and the references therein Langevin equation is widely used to describe the evolution of physical phenomena in fluctuating environments However, for systems in complex media, ordinary Langevin equation does not provide the correct description of the dynamics One of the possible generalizations of Langevin equation is to replace the ordinary derivative by a fractional derivative in it This gives rise to fractional Langevin equation, see for instance 9–12 and the references therein In this paper, we consider the following boundary value problem of Langevin equation with two different fractional orders: C Dβ C Dα u0 λ ut −u T , f t, u t u u T t∈ 0, T , 1.1 0, Boundary Value Problems where T is a positive constant, < α ≤ 2, < β ≤ 1, C Dα , and C Dβ are the Caputo fractional derivatives, f : 0, T × R → R is continuous, and λ is a real number The organization of this paper is as follows In Section 2, we recall some definitions of fractional integral and derivative and preliminary results which will be used in this paper In Section 3, we will consider the existence results for problem 1.1 In Section 4, we will give an example to ensure our main results Preliminaries In this section, we present some basic notations, definitions, and preliminary results which will be used throughout this paper Definition 2.1 The Caputo fractional derivative of order α of a function f : 0, ∞ → R, is defined as C Dα f t t Γ n−α t−s n−α−1 f n n − < α < n, n s ds, α 1, 2.1 where α denotes the integer part of the real number α Definition 2.2 The Riemann-Liouville fractional integral of order α > of a function f t , t > 0, is defined as t Γα I αf t t−s α−1 f s ds, 2.2 provided that the right side is pointwise defined on 0, ∞ Definition 2.3 The Riemann-Liouville fractional derivative of order α > of a continuous function f : 0, ∞ → R is given by Dα f t d dt Γ n−α n t t−s n−α−1 f s ds, 2.3 where n α and α denotes the integer part of real number α, provided that the right side is pointwise defined on 0, ∞ Lemma 2.4 see Let α > 0, then the fractional differential equation ut where ci ∈ R, i c0 0, 1, 2, , n − 1, n c1 t α ··· c1 t c2 t2 c0 Dα u t cn−1 tn−1 , has solution 2.4 u t c2 t2 C Lemma 2.5 see Let α > 0, then I αC Dα u t for some ci ∈ R, i 0, 1, 2, , n − 1, n α ··· cn−1 tn−1 , 2.5 Boundary Value Problems Lemma 2.6 The unique solution of the following boundary value problem C Dβ C Dα λ ut t ∈ 0, T , < α ≤ 2, < β ≤ 1, y t, 2.6 −u T , u0 u u T 0, is given by t − s α−1 Γ α t ut − T s T −s Γα T T α − 2tα 2αT α−1 β−1 s−τ Γ β α−1 s y τ dτ − λu s s−τ Γ β T − s α−2 Γ α−1 s β−1 ds y τ dτ − λu s s−τ Γ β β−1 ds 2.7 y τ dτ − λu s ds Proof Similar to the discussion of 9, equation 1.5 , the general solution of C Dβ C Dα λ ut y t 2.8 can be written as t ut t − s α−1 Γ α s s − τ β−1 y τ dτ − λu s Γ β By the boundary conditions u c0 Γ α T αT α−1 c1 0 and u T − s α−2 Γ α−1 s s−τ Γ β u T β−1 c0 tα − c1 t − c2 Γα 2.9 0, we obtain y τ dτ − λu s ds, 0, c2 uT ds − − T T 2α T − s α−1 Γ α T T − s α−2 Γ α−1 s s − τ β−1 y τ dτ − λu s Γ β s 2.10 ds s − τ β−1 y τ dτ − λu s Γ β ds 4 Boundary Value Problems Hence, t − s α−1 Γ α t ut − T s T −s Γα α−1 s T − s α−2 Γ α−1 y τ dτ − λu s s−τ Γ β T T α − 2tα 2αT α−1 β−1 s−τ Γ β s β−1 ds y τ dτ − λu s s−τ Γ β β−1 ds 2.11 y τ dτ − λu s ds Lemma 2.7 Krasnoselskii’s fixed point theorem Let E be a bounded closed convex subset of a Banach space X, and let S, T be the operators such that i Su Tv ∈ E whenever u, v ∈ E, ii S is completely continuous, iii T is a contraction mapping Then there exists z ∈ E such that z Sz Tz Lemma 2.8 Holder inequality Let p > 1, 1/p ă the following inequality holds: b 1/p b f x g x dx ≤ f x a 1, f ∈ Lp a, b , g ∈ Lq a, b , then 1/q p b dx a 1/q |g x |q dx 2.12 a Main Result In this section, our aim is to discuss the existence and uniqueness of solutions to the problem 1.1 Let Ω be a Banach space of all continuous functions from 0, T → R with the norm u supt∈ 0,T {|u t |} Theorem 3.1 Assume that (H1) there exists a real-valued function μ t ∈ L1/γ 0, T , R f t, u − f t, v ≤ μ t |u − v|, for some γ ∈ 0, such that for almost all t ∈ 0, T , u, v ∈ R 3.1 If Λ¸ 4α β−γ Γ β−γ 2αΓ β Γ α μ∗ T α β−γ where γ ∈ 0, , β / γ, < α ≤ 2, < β ≤ 1, μ∗ unique solution β−γ 1−γ β−γ T μτ 1−γ 1/γ 2|λ|T α < 1, Γα γ 3.2 dτ , then problem 1.1 has a Boundary Value Problems Proof Define an operator F : Ω → Ω by t − s α−1 Γ α t Fu t 0 T − s α−1 Γ α T − s − τ β−1 f τ, u τ dτ − λu s Γ β s T α − 2tα 2αT α−1 T s s−τ Γ β T − s α−2 Γ α−1 s β−1 ds f τ, u τ dτ − λu s s−τ Γ β β−1 3.3 ds f τ, u τ dτ − λu s ds supt∈ 0,T |f t, | and choose Let M 1−δ 4α β MT α 2αΓ α β where δ is such that Λ ≤ δ < Now we show that FBr ⊂ Br , where Br inequality, we have β ≤ r, 3.4 {u ∈ Ω : u r} For u Br , by Holder ă | Fu t | t − s α−1 Γα t − T t T T T T − s α−2 Γ α−1 s T −s Γα s α−1 s f τ, u τ s s − τ β−1 Γ β s s−τ Γ β s−τ Γ β β−1 ds s − τ β−1 f τ, u τ dτ − λu s Γ β β−1 s−τ Γ β s ds s − τ β−1 f τ, u τ dτ − λu s Γ β T − s α−2 Γ α−1 t − s α−1 Γα t s T − s α−1 Γα T 2α ≤ T t − s α−1 Γα 0 T − s α−1 Γα T α − 2tα 2αT α−1 ≤ s − τ β−1 f τ, u τ dτ − λu s Γ β s − f τ, f τ, u τ − f τ, β−1 f τ, s − τ β−1 μ τ |u τ | Γ β dτ f τ, dτ f τ, − f τ, f τ, u τ μ τ |u τ | f τ, ds f τ, |λu s | ds dτ dτ |λu s | ds dτ |λu s | ds |λu s | ds |λu s | ds Boundary Value Problems ≤ u T − s α−2 Γ α−1 T T 2α 0 0 ≤ T u Γα Γ β t−s u 2Γ α Γ β M 2Γ α Γ β s α−1 t s−τ T β−1 T −s Γα t−s T T −s 1−γ β−γ α−1 ds T − s α−2 ds Γ α−1 T 1−γ 1/ 1−γ s s ds s α−1 1−γ 1/ 1−γ β−1 s−τ s ds s α−2 s μτ dτ s−τ β−1 1−γ t T −s t−s T s ds α−1 β−γ s M Γα Γ β ds T −s α−1 β−γ s ds Tμ∗ u 2αΓ α − Γ β 1−γ β−γ TM 2αΓ α − Γ β T 1−γ 1−γ ds T T −s α−2 β γ s dτ 1/γ μ τ α−2 β−γ s s ds t M 2Γ α Γ β ds T −s γ dτ dτ |λ|T α u 2Γ α α−2 β 1−γ 1/γ 0 1/ 1−γ ds |λ|T α u 2Γ α α−1 β T −s T γ dτ 0 1−γ β−γ 1/γ μτ dτ T ds |λ|T α u Γ α α−1 β T −s 0 T s − τ β−1 dτ ds Γ β s ds T|λ| u 2α dτ ds TM 2αΓ α − Γ β μ∗ u 2Γ α Γ β β−1 t − s α−1 Γα s − τ β−1 μ τ dτ Γ β s |λ| u dτ ds |λu s | ds dτ T u 2αΓ α − Γ β μ∗ u Γα Γ β M β−1 s−τ Γ β t ds s − τ β−1 μ τ dτ Γ β s s f τ, T − s α−1 Γ α s−τ Γ β M Γα Γ β ≤ s T − s α−2 Γ α−1 t μ τ |u τ | T u T − s α−2 Γ α−1 T T u 2α TM 2α T − s α−1 Γα T β−1 s − τ β−1 μ τ dτ Γ β s t − s α−1 ds Γ α t |λ| u s−τ Γ β t − s α−1 Γ α t M s 2|λ|T α u Γα t−s α−1 β s ds T T −s α−1 β s ds ds Boundary Value Problems μ∗ u tα β−γ Γα Γ β 1−γ β−γ μ∗ u T α β−γ 2Γ α Γ β 1−γ 1−γ β−γ 1−γ MT α β 2αΓ α − Γ β rμ∗ T α β−γ 2Γ α Γ β 1−γ β−γ 1 α−1 β−γ 1−η 1−γ η 1−η η 1−η 1−ξ α−1 β−γ 1−γ β−γ 1 1−η α−1 β η dη dη η dη ξ MT α Γ αΓ β dξ α−1 β−γ 1−η η 1−γ 1−η α−2 β−γ η β 1−ξ α−1 β ξ dξ MT α β 2Γ α Γ β dη MT α β 2αΓ α − Γ β ξ dξ 2|λ|T α u Γα α−2 β 1−γ rμ∗ T α β−γ 2αΓ α − Γ β α−1 β 0 1−γ β−γ 1−ξ MT α β 2Γ α Γ β dη α−2 β−γ 1−γ Mtα β Γα Γ β dξ 1−γ β−γ rμ∗ T α β−γ ΓαΓ β ξ μ∗ u T α β−γ 2αΓ α − Γ β ≤ α−1 β−γ 1−ξ 1−η α−1 β η dη dη 1−η 2|λ|T α r Γ α α−2 β η dη 3.5 Take notice of Beta functions: B β−γ 1, α 1−ξ α−1 β−γ ξ dξ B β 1, α B β−γ 1, α − 1−ξ α−1 β ξ dξ 1−η 1, α − 1 η Γ α Γ β−γ 1−η α−2 β−γ 1−η α−2 β η Γ α β−γ Γ α−1 Γ β η dη Γ α β 1 1 β rμ∗ Γ β − γ Γ β Γ α Tα β−γ rμ∗ Γ β − γ 2Γ β Γ α 1−γ β−γ β−γ 1 Tα β−γ β−γ 1−γ β−γ 1−γ 1−γ , , 3.6 , We can get | Fu t | ≤ β−γ Γ α η dη Γ α−1 Γ β−γ dη Γ α dη Γα Γ β α−1 β 0 B β α−1 β−γ 0 1−η MT α β Γ α β MT α β 2Γ α β Boundary Value Problems rμ∗ Γ β − γ Tα 2αΓ β Γ α β−γ β−γ Γ β−γ 4α 2αΓ β Γ α β MT α 4α 2αΓ α ≤ Λ β 1−γ β−γ β−γ μ∗ T α β−γ 1−γ MT α β 2αΓ α β 1−γ β−γ β−γ 1−γ 2|λ|T α r Γα 2|λ|T α Γα r β 1−δ r ≤ r 3.7 Therefore, Fu t ≤ r For u, v ∈ Ω and for each t ∈ 0, T , based on Holder inequality, we obtain ă | Fu t Fv t | ≤ t − s α−1 Γα t t |λ| T T 2α |λ|T 2α ≤ T −s Γα |λ| s−τ Γ β β−1 − f τ, v τ f τ, u τ dτ ds − f τ, v τ dτ t − s α−1 |u s − v s |ds Γα s T α−1 s − τ β−1 f τ, u τ Γ β s ds T − s α−1 |u s − v s |ds Γ α T − s α−2 Γ α−1 T T u−v Γα Γ β s s−τ Γ β β−1 f τ, u τ − f τ, v τ dτ ds T − s α−2 |u s − v s |ds Γ α−1 t t−s s α−1 u−v 2Γ α Γ β s−τ β−1 μ τ dτ ds T T −s α−1 T u−v 2αΓ α − Γ β s s−τ β−1 μ τ dτ ds T T −s α−2 s |λ|T α u−v Γα s−τ β−1 |λ|T α 2Γ α μ τ dτ ds u−v |λ|T α 2Γ α u−v Boundary Value Problems ≤ u−v Γα Γ β t t−s 2|λ|T α Γα μ∗ u − v ΓαΓ β T T −s 1−γ s μ τ dτ s α−1 1−γ 1/ 1−γ β−1 s−τ T 1/γ γ dτ s μτ dτ ds s α−2 T −s 1/γ γ dτ ds 0 s−τ β−1 1/ 1−γ 1−γ s dτ μτ 1/γ γ dτ ds u−v 1−γ β−γ 1−γ 1−γ β−γ t−s 1−γ T 1−γ β−γ β−γ μ∗ u − v T α β−γ 2αΓ α − Γ β s T −s 1−γ T ds α−1 β−γ s ds T −s α−2 β−γ s ds 1−γ 1−ξ α−1 β−γ ξ 2|λ|T α u−v Γα dξ 1−γ β−γ 1−γ 1−γ β−γ 1−γ β−γ Γ β−γ 2αΓ β Γ α α−1 β−γ 1−γ β−γ β−γ μ∗ u − v T α 2Γ α Γ β t μ∗ T u − v 2αΓ α − Γ β 4α 1/ 1−γ β−1 0 μ∗ u − v 2Γ α Γ β μ∗ u − v tα Γ αΓ β s−τ T u−v 2αΓ α − Γ β ≤ s α−1 u−v 2Γ α Γ β ≤ 1−η α−1 β−γ η dη 1−η α−2 β−γ dη η μ∗ T α β−γ 1 β−γ 1−γ β−γ 1−γ 2|λ|T α Γα 2|λ|T α Γ α u−v u−v Λ u−v 3.8 Since Λ < 1, consequently F is a contraction As a consequence of Banach fixed point theorem, we deduce that F has a fixed point which is a solution of problem 1.1 Corollary 3.2 Assume that (H1) There exists a constant L > such that f t, u − f t, v ≤ L|u − v|, ∀t ∈ 0, T , u, v ∈ R 3.9 10 Boundary Value Problems If β LT α 4α 2αΓ α β β 2|λ|T α < 1, Γα 1 3.10 then problem 1.1 has a unique solution Theorem 3.3 Suppose that (H1) and the following condition hold: (H2) There exists a constant l ∈ 0, and a real-valued function m t ∈ L1/l 0, T , R that ≤m t , f t, u for almost every t ∈ 0, T , u ∈ R such 3.11 Then the problem 1.1 has at least one solution on 0, T if μ∗ T α β−γ Γ β−γ 2α 2αΓ β Γ α β−γ 1−γ β−γ β−γ 1−γ |λ|T α < Γα 3.12 Proof Let us fix β−l Γ β−l 4α 2αΓ β Γ α β−l m∗ T α β−l 1 − 2|λ|T α / Γ α 1−l β−l 1−l ≤ r; 3.13 l T 1/l here, m∗ dτ ; consider Br {u ∈ Ω : u ≤ r}, then Br is a closed, bounded, m τ and convex subset of Banach space Ω We define the operators S and T on Br as t − s α−1 Γ α t Su t Tu t − T s α−1 T −s Γα s T T α − 2tα 2αT α−1 s − τ β−1 f τ, u τ dτ − λu s Γ β β−1 s−τ Γ β T − s α−2 Γ α−1 s ds, f τ, u τ dτ − λu s s−τ Γ β β−1 f τ, u τ dτ − λu s For u, v ∈ Br , based on Holder inequality, we find that ă |Su Tv| t t s Γα T s T − s α−1 Γα s − τ β−1 f τ, u τ Γ β s dτ s − τ β−1 f τ, v τ Γ β |λu s | ds dτ ds |λv s | ds 3.14 ds Boundary Value Problems T 2α ≤ T T − s α−2 Γ α−1 t Γα Γ β s T ≤ m∗ Γα Γ β m∗ 2Γ α Γ β m∗ T α β−l Γα Γ β m∗ T α β−l 2Γ α Γ β T s−τ T −s s α−2 β−1 m τ dτ ds s−τ β−1 s s−τ β−1 1−l 1/ 1−l s T −s 1−l β−l 1−l T − s α−2 ds Γ α−1 1/l dτ ds β−1 1/ 1−l 1−l l s dτ m τ s α−2 t−s 1/l ds dτ T |λ| u 1−l s−τ β−1 1/ 1−l 1−l m τ α−1 β−l s 1/l dτ ds T − s α−1 ds Γ α l s dτ |λ|T u 2α T T − s α−2 ds Γ α−1 ds T T −s α−1 β−l s ds 1−l 1−l β−l 1−l T T −s α−2 β−l s ds 1−ξ α−1 β−l ξ 2|λ|T α r Γα dξ 1−l 1−l β−l 1−η α−1 β−l η dη 1−l 1−l β−l β−l Γ β−l 2αΓ β Γ α ds l m τ t α−1 1−l β−l 1−l β−l s−τ t − s α−1 ds Γα T |λ|T u 2α T s dτ T −s Γα α−1 T −s T |λ| u m τ dτ ds T t − s α−1 ds Γ α t |λ| u α−1 t−s m∗ T α β−l 2αΓ α − Γ β 4α m τ dτ ds |λv s | ds m∗ T 2αΓ α − Γ β ≤ s α−1 T −s T 2αΓ α − Γ β t dτ 0 2Γ α Γ β β−1 s−τ t Γα Γ β f τ, v τ T 2αΓ α − Γ β |λ| u s α−1 t−s β−1 s−τ Γ β 2Γ α Γ β ≤ 11 1−η α−2 β−l η m∗ T α β−l β−l 1−l β−l 1−l dη 2|λ|T α r Γα 2|λ|T α r Γα ≤ r 3.15 Thus, Su Tv ≤ r, so Su Tv ∈ Br 12 Boundary Value Problems For u, v ∈ Ω and for each t ∈ 0, T , by the analogous argument to the proof of Theorem 3.1, we obtain | Tu t − Tv t | ≤ T |λ| T 2α |λ|T 2α ≤ T −s Γα α−1 s β−1 − f τ, v τ f τ, u τ dτ ds T − s α−1 |u s − v s |ds Γ α T T − s α−2 Γ α−1 T T s β−1 s−τ Γ β − f τ, v τ f τ, u τ dτ ds T − s α−2 |u s − v s |ds Γ α−1 u−v 2Γ α Γ β T u−v 2Γ α Γ β s α−1 T −s T β−1 s−τ μ τ dτ ds T u−v 2αΓ α − Γ β ≤ s−τ Γ β T β−1 s−τ |λ|T α u−v 2Γ α μ τ dτ ds s α−1 T −s T u−v 2αΓ α − Γ β s α−2 T −s |λ|T α u−v 2Γ α β−1 1/ 1−γ s−τ 1−γ μ τ s dτ ds s α−2 T −s 1/γ 1−γ T γ s dτ s−τ β−1 1/ 1−γ dτ γ μ τ 1/γ dτ ds |λ|T α u−v Γα ≤ μ∗ u − v 2Γ α Γ β 1−γ β−γ μ∗ T u − v 2αΓ α − Γ β μ∗ u − v T α 2Γ α Γ β β−γ μ∗ u − v T α β−γ 2αΓ α − Γ β ≤ 2α 1−γ α−1 β−γ T −s s ds 1−γ β−γ 1−γ β−γ 1−γ T T −s α−2 β−γ s ds 1−γ 1−η α−1 β−γ η |λ|T α Γ α u−v dη 1−γ β−γ β−γ Γ β−γ 2αΓ β Γ α T 1−γ 1−η α−2 β−γ η μ∗ T α β−γ 1 β−γ 1−γ β−γ 1−γ dη |λ|T α u−v Γα |λ|T α Γ α u−v 3.16 Boundary Value Problems 13 From the assumption β−γ Γ β−γ 2α 2αΓ β Γ α μ∗ T α β−γ β−γ 1−γ β−γ 1−γ |λ|T α < 1, Γα 3.17 it follows that T is a contraction mapping The continuity of f implies that the operator S is continuous Also, S is uniformly bounded on Br as Γ β−l Su ≤ On the other hand, let N σ m∗ T α β−l β−l Γ β Γ α max t,u ∈ 0,T ⎧ ⎨1 ×Br |f εΓ α β 2N ⎩2 1−l β−l t, u t | 1/ α β 1−l |λ|T α r Γ α 3.18 1, for all ε > 0, setting ⎫ εΓ α , 2|λ|r 1/α ⎬ ⎭ 3.19 For each u ∈ Br , we will prove that if t1 , t2 ∈ 0, T and < t2 − t1 < σ, then | Su t2 − Su t1 | < ε 3.20 In fact, we have | Su t2 − Su t1 | t2 t2 − s Γα t1 − t1 t2 t1 t1 − t1 t1 s α−1 α−1 t1 − s Γα α−1 s s α−1 t2 − s Γα s − τ β−1 f τ, u τ dτ − λu s Γ β s − τ β−1 f τ, u τ dτ − λu s Γ β s 0 t2 α−1 t2 − s Γα t2 − s s − τ β−1 f τ, u τ dτ − λu s Γ β s t1 − s Γα t2 − s Γα α−1 s − τ β−1 f τ, u τ dτ − λu s Γ β s−τ Γ β − t1 − s Γα α−1 s α−1 s−τ Γ β β−1 f τ, u τ dτ − λu s s β−1 ds ds ds ds ds s − τ β−1 f τ, u τ dτ − λu s Γ β f τ, u τ dτ − λu s ds ds 14 Boundary Value Problems ≤ t1 t2 − s α−1 |λ| u t1 t2 − s t1 ≤ Γ α t2 − s Γα α−1 s N β − t1 − s Γ α t2 s − τ β−1 f τ, u τ Γ β ds ds |λ|r Γα α β − t1 dτ α−1 s − τ β−1 f τ, u τ Γ β α β s α−1 t2 α−1 − t1 − s Γα dτ ds t2 |λ| u t1 t2 − s Γ α α−1 ds tα − tα 3.21 In the following, the proof is divided into two cases Case For σ ≤ t1 < t2 < T, we have | Su t2 − Su t1 | ≤ ≤ Γ α N β N Γ α β < α β − t1 α β σα N σα Γ α β < α β |λ|r Γα α β t2 |λ|r ασ α−1 t2 − t1 Γα t2 − t1 |λ|r α σ Γ α β ε β−1 tα − tα α ε < ε 3.22 Case for ≤ t1 < σ, t2 < 2σ, we have | Su t2 − Su t1 | ≤ ≤ N Γ α β N β Γ α < N Γ α β < ε α β t2 α β t2 2σ α β − t1 |λ|r Γα tα − tα |λ|r α t Γ α α β |λ|r 2σ Γ α α 3.23 ε ε Therefore, S is equicontinuous and the Arzela-Ascoli theorem implies that S is compact on Br , so the operator S is completely continuous Boundary Value Problems 15 Thus, all the assumptions of Lemma 2.7 are satisfied and the conclusion of Lemma 2.7 implies that the boundary value problem 1.1 has at least one solution on 0, T Corollary 3.4 Suppose that the condition (H1) hold and, assume that β LT α 2α 2αΓ α β β |λ|T α < Γα 1 3.24 Further assume that (H2) there exists a constant K > such that ≤ K, f t, u ∀t ∈ 0, T , u ∈ R, 3.25 then problem 1.1 has at least one solution on 0, T Example Let α 2, β 1, λ 1/8, T C π/2 We consider the following boundary value problem D1 C ut D2 π u u0 f t, u t , ≤ t ≤ 0, u π u π , 4.1 0, where f t, u t u 2 u , t, u ∈ 0, T × 0, ∞ 4.2 Because of |f t, u − f t, v | ≤ 1/4 |u − v|, let μ t ≡ 1/4, then μ t ∈ L2 0, π/2 , we have √ 1/2 γ √ T π/2 μ τ 1/γ dτ 1/4 dτ π/4 Further, γ 1/2 and μ∗ 0 4α β−γ Γ β−γ 2αΓ β Γ α μ∗ T α β−γ 17/2 Γ 3/2 μ∗ T 5/2 4Γ Γ 7/2 17π 15 × 64 β−γ 1−γ β−γ 1−γ 2|λ|T α Γ α 2|λ|T Γ π2 32 ≈ 0.86 < Then BVP 4.1 has a unique solution on 0, π/2 according to Theorem 3.1 4.3 16 Boundary Value Problems On the other hand, we find that β−γ Γ β−γ 2α 2αΓ β Γ α β−γ 9/2 Γ 3/2 μ∗ T 5/2 4Γ 7/2 9π 64 × 15 μ∗ T α β−γ 1−γ β−γ 1−γ |λ|T α Γ α |λ|T Γ3 4.4 π2 64 ≈ 0.44 < Then BVP 4.1 has at least one solution on 0, π/2 according to Theorem 3.3 Acknowledgments This work was supported by the Natural Science Foundation of China 10971173 , the Natural Science Foundation of Hunan Province 10JJ3096 , the Aid Program for Science and Technology Innovative Research Team in Higher Educational Institutions of Hunan Province, and the Construct Program of the Key Discipline in Hunan Province References A Alsaedi, “Existence of solutions for integrodifferential equations of fractional order with antiperiodic boundary 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